0107. Binary Tree Level Order Traversal II (E)

Binary Tree Level Order Traversal II (E)

题目

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

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题意

按从下到上的顺序记录二叉树的每一层。

思路

BFS后再将结果集逆序。

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代码实现

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List> levelOrderBottom(TreeNode root) {
        List> ans = new ArrayList<>();
        Queue q = new ArrayDeque<>();

        if (root != null) {
            q.offer(root);
        }

        while (!q.isEmpty()) {
            List list = new ArrayList<>();
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = q.poll();
                list.add(cur.val);
                if (cur.left!=null) q.offer(cur.left);
                if (cur.right!=null) q.offer(cur.right);
            }
            ans.add(list);
        }
        
        Collections.reverse(ans);
        return ans;
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrderBottom = function (root) {
  let bfs = []
  let q = []
  if (root) {
    q.push(root)
  }
  while (q.length > 0) {
    let size = q.length
    let tmp = []
    while (size > 0) {
      let p = q.shift()
      tmp.push(p.val)
      if (p.left) q.push(p.left)
      if (p.right) q.push(p.right)
      size--
    }
    bfs.push(tmp)
  }
  return bfs.reverse()
}